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...like that, so the pressure on bitter taste alleles is reduced. PTC non-tasters do still taste many bitter compounds, just not certain ...

A. Since those who cannot taste PTC are homozygous tt, the frequency of the tt genotype (q^2) = 65/215 = 0...

The ability to taste or not to taste is genetic. Tasting is a dominant phenotype but seems to be variable in its expressivity: some tasters are "supertasters," and are assumed to be those individuals who...

If you're crossing BbTt x BbTt, the proportion of bbT_ would be 3/16. When doing a dihybrid cross, 9/16 are dominant for both traits, 3/16 are dominant for one and recessive for the other, 3/16 are the other way around, and 1/16 is...

Since 37/125 could not taste PTC , fr(tt) must be 37/125 = 0.2960. Therefore, fr(t) = sqrt...

...0.4 * 0.6 * 1092 = 524 (approximately) (answer) C) How many people can taste PTC ? 64% * 1092 = 699 (approx) (answer) Check: This should be the same as (p...

Okay. You have 150 tasters and 215-150 = 65 nontasters. q^2 = 65/215 = .302 q = 0.55 freq(t) = q = 0.55 (answer) freq(T) = p = 1-q = 0.45 (answer) Check that: p^2 + 2pq + q^2 = 1 .45^2 + 2 * .45*.55 + .55^2 = 1 check 215 * (.45^2 + 2 * .45*.55) = 150 check 215 * (.55^2) = 65 check

Since it is a dominant trait you must be heterozygous. So if P=taster, your mother is pp, father is Pp (because your brother is a non-taster so pp) and so you are Pp.